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swat32
1 year ago
13

Find the volume of a pyramid with a square base, where the side length of the base is

Mathematics
1 answer:
Bumek [7]1 year ago
4 0

Answer:

1025.76

Step-by-step explanation:

Formula for a right rectangular pyramid is (l*w*h)/3

The length and width are both 18.7, so we can plug that in. The height is 8.8.

It becomes:

(18.7*18.7*8.8)/3, which equals the answer above.

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Evaluate 8 − 12 ÷ 2 + 3. (1 point)
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Answer:

Using PEMDAS, you should get the answer of 5.

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2 years ago
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Thanks to all those who support my question!
Mazyrski [523]

Answer:

no, because intergers are negative. they will never be able to be greater

Step-by-step explanation:

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2 years ago
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Help!<br> What would d be?
Oxana [17]

Answer:

d = 17.5499287748 inches or 17.5 inches to 1 decimal place

Step-by-step explanation:

find the diagonal line passing through the middle of the bottom rectangle:

a² + b² = c²

10² + 8² = c²

100 + 64 = 164

c² = 164

c = √164

c = 12.8062484729

find d:

a² + b² = c²

12² + 12.80624847298² = c²

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c = 17.5499287748

therefore d = 17.5499287748 inches or 17.5 inches to 1 decimal place

6 0
3 years ago
I dont know how to do graphing pls help
yarga [219]

Answer:

Solution: (-1, -1)

Step-by-step explanation:

y=4x+2

y=-4/3x-2

Solve by graphing.

First, you need to plot the y-intercept.

y=4x+<u>2</u>

2 will be your y-intercept.

Next, you plot your slope.

y=<u>4x</u>+2

From your y-intercept, you will go up 4 and right one space, if you run out of space go down 4 and left 1.

Now repeat the same steps for the next one.

y=-4/3x<u>-2</u>

Plot your y-intercept.

y=<u>-4/3x</u>-2

because your slope is negative you will go down 4 and right 3, if you run out of room go up 4 and left 3.

Then draw connecting lines and wherever the lines intersect, that's going to be your solution. In this case, the solution is (-1, -1).

Hope this helps :)

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2 years ago
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Which is a true statement concerning outliers for the data set summarized by the box-and-whisker plot shown?
Ray Of Light [21]
B :^) hope this helps!
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